# Horizontal Projectile motion

• NemoMnemosyne

#### NemoMnemosyne

This was an even numbered problem from the book so there are no answers to make sure I've done this correctly or not.

## Homework Statement

An arrow fires horizontally at 41 m/s travels 23 m horizontally. From what height was it fired?

Vi = 41 m/s
a = g = -9.8m/s
x = 23m

## Homework Equations

Vf = Vi + at
X = Xi + Vit + 1/2at^2
Y = Yi + Vit + 1/2at^2

## The Attempt at a Solution

A.) Time to reach ground

Vf = Vi +at =>
Vf - Vi/a = t =>
0 - 41/-9.8 = 4.2s

B.) Height of Cliff
y = Vit + 1/2at^2 =>
y = 41(4.2) + 1/2(-9.8)(4.2^2) =>
y = 85.76m

Hi!
It appears you have got yourself into a bit of a muddle. The real trick with questions like this is to separate out all the x and y parameters (speed (initial, u, and final, v), acceleration (a), distance (s), time (t)).

So, in x, we have:

Vx = Ux = 41m/s
ax = 0 (nothing to accelerate the particle in the x plane)
sx = 23m
tx = ?

And in y,

Uy = 0
Vy = ?
ay = g = 9.8m/s/s
sy = what we want to find
ty = tx (can you see why?)

Now, clearly you want to know how long the arrow is in the air for. You haven't got enough information from the y variables, but you have plenty in x! It turns out you just need to do a simple speed = distance over time calculation.

See what you can do with this new info.

You have the right set of equations but you are getting x and y directions mixed up. The arrow is fired horizontally and that horizontal velocity is unchanging. The initial velocity in the vertical direction is zero and changes constantly until impact.